Find an equation of.

the plane that passes through the line of intersection of the planes 

*x* − *z* = 3 and *y* + 2*z* = 2

 and is perpendicular to the plane 

*x* + *y* − 4*z* = 5

To find the equation of the plane that passes through the line of intersection of two given planes and is perpendicular to a third plane, we can follow a systematic approach:

### Step-by-Step Approach:

1. **Given Planes:**
   - Plane 1: \( x - z = 3 \)
   - Plane 2: \( y + 2z = 2 \)
   - Plane 3 (the perpendicular plane): \( x + y - 4z = 5 \)

   We seek the equation of a plane that passes through the intersection of Plane 1 and Plane 2 and is perpendicular to Plane 3.

2. **Line of Intersection of Two Planes (Parametric Form):**
   The general form of a plane passing through the line of intersection of two planes can be expressed as:
   \[
   \Pi : \Pi_1 + \lambda \Pi_2 = 0
   \]
   where \( \Pi_1 \) and \( \Pi_2 \) are the equations of the two given planes, and \( \lambda \) is a scalar constant. 

   Substituting the equations of the two planes:
   \[
   (x - z - 3) + \lambda (y + 2z - 2) = 0
   \]
   Expanding this gives the equation of the family of planes:
   \[
   x - z - 3 + \lambda y + 2\lambda z - 2\lambda = 0
   \]
   Simplifying:
   \[
   x + \lambda y + (-1 + 2\lambda)z = 3 + 2\lambda
   \]

   This is the general form of a plane that passes through the line of intersection of the two given planes.

3. **Perpendicularity Condition:**
   The plane we seek is also perpendicular to the plane \( x + y - 4z = 5 \). For two planes to be perpendicular, the dot product of their normal vectors must be zero.

   - The normal vector to the plane \( x + \lambda y + (-1 + 2\lambda)z = 3 + 2\lambda \) is \( \mathbf{n}_1 = (1, \lambda, -1 + 2\lambda) \).
   - The normal vector to the plane \( x + y - 4z = 5 \) is \( \mathbf{n}_2 = (1, 1, -4) \).

   The dot product of \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \) must be zero:
   \[
   (1)(1) + (\lambda)(1) + (-1 + 2\lambda)(-4) = 0
   \]
   Expanding:
   \[
   1 + \lambda + 4 - 8\lambda = 0
   \]
   Simplifying:
   \[
   5 - 7\lambda = 0
   \]
   Solving for \( \lambda \):
   \[
   \lambda = \frac{5}{7}
   \]

4. **Final Equation of the Plane:**
   Substituting \( \lambda = \frac{5}{7} \) into the general equation of the family of planes:
   \[
   x + \frac{5}{7} y + \left(-1 + 2 \times \frac{5}{7}\right)z = 3 + 2 \times \frac{5}{7}
   \]
   Simplifying:
   \[
   x + \frac{5}{7} y + \frac{3}{7} z = \frac{31}{7}
   \]
   Multiplying through by 7 to eliminate fractions:
   \[
   7x + 5y + 3z = 31
   \]

Thus, the equation of the desired plane is:
\[
7x + 5y + 3z = 31
\]

Would you like further details on any step? Here are some related questions:

1. How do you find the intersection line of two planes geometrically?
2. What is the geometric meaning of the dot product between two normal vectors?
3. Can two planes be both perpendicular and intersect along a line?
4. How would you verify if two planes are parallel or coincident?
5. What is the significance of the scalar parameter \( \lambda \) in defining the family of planes?

**Tip:** The normal vector of a plane reveals much about its orientation in space, so understanding its components helps solve problems involving perpendicularity and intersections.

Find an equation of the plane that passes through the line of intersection of the planes x − z = 3 and y + 2z = 2 and is perpendicular to the plane x + y − 4z = 5.

Learn how to find the equation of a plane passing through the intersection of the planes x − z = 3 and y + 2z = 2, and perpendicular to the plane x + y − 4z = 5. This problem involves key concepts such as dot products, parametric equations, and conditions for perpendicular planes. Suitable for advanced high school or undergraduate students.

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